Young boy Artem tries to paint a picture, and he asks his mother Medina to help him. Medina is very busy, that’s why she asked for your help.
Artem wants to paint an n×m board. Each cell of the board should be colored in white or black.
Lets B be the number of black cells that have at least one white neighbor adjacent by the side. Let W be the number of white cells that have at least one black neighbor adjacent by the side. A coloring is called good if B=W+1.
The first coloring shown below has B=5 and W=4 (all cells have at least one neighbor with the opposite color). However, the second coloring is not good as it has B=4, W=4 (only the bottom right cell doesn’t have a neighbor with the opposite color).
Please, help Medina to find any good coloring. It’s guaranteed that under given constraints the solution always exists. If there are several solutions, output any of them.
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1≤t≤20). Each of the next t lines contains two integers n,m (2≤n,m≤100) — the number of rows and the number of columns in the grid.
Output
For each test case print n lines, each of length m, where i-th line is the i-th row of your colored matrix (cell labeled with ‘B’ means that the cell is black, and ‘W’ means white). Do not use quotes.
It’s guaranteed that under given constraints the solution always exists.
Example
Input
2
3 2
3 3
Output
BW
WB
BB
BWB
BWW
BWB
Note
In the first testcase, B=3, W=2.
In the second testcase, B=5, W=4. You can see the coloring in the statement.
思路：构造出任意的一种就可以了。那么我们就构造W=1，B=2的，把左上角弄成W，其余的全是B就可以了。
代码如下：
#include<bits/stdc++.h>
#define ll long long
using namespace std;
int a,b;
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&a,&b);
for(int i=1;i<=a;i++)
for(int j=1;j<=b;j++)
{
if(i==1&&j==1) cout<<'W';
else cout<<'B';
if(j==b) cout<<endl;
}
}
return 0;
}
努力加油a啊，(o)/~
